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Answer 8:
Answer 8:
(a) Consider a self-adjoint operator
3
()
PMR
Î
such that
P(1, 2, 3) = P(0, 0, 0) and P(2;5;7) = P(2;5;7).
Every self-adjoint operator has orthogonal eigen-vectors. However,
(1,2,3) is an eigen vector for P with eigen value 0 and (2, 5, 7) is an
eigen-vector for P with eigen value 1. However, (1, 2, 3) and (2, 5, 7) are
not orthogonal, thus P cannot be self-adjoint.
(b) A linear transformation T : V →V, dim(V)= n , when we can find an
ordered basis B of V such that is diagonal.
there exists a diagonal matrix D such that A = P-1DP. Then we have 
Let 
For B ={1, x} for ,
 The characteristic polynomial of T is given by 
 
        Thus, T has two eigenvalues λ = 1, 2. Further, for the eigenvalues λ = 1
 
Thus (  - I ) X = 0 implies
 
   Thus, X =  is an eigenvector for  for λ = 1.
(c) V is a direct sum of null (P) and range (P) if the intersection of null
(P) and range (P) is just zero and every vector in V can be
written as a sum of a vector in null (P) with a vector in range (P).
Suppose x is in both null(P) and range(P). Because x is in null(P),
P(x) = 0. But x is also in range(P) so there is a y in V such that
P y = x. Since
2
2
2
,
,
,
()()()
PP
PyPy
therefore
PyPyPx
=
=
==
2
()0,
()(){0}
PyPyPx
thisgives
nullPrangeP
===
Ç=
Let v be a vector in V . Then v = P v + (v - P v). So every vector in V
can be written as a sum of a vector in null(P) with a vector in range(P).
Answer 9: If A is an m×n matrix and Ax = b always has at least one
solution for any choice of b ∈ Cm, that means that any vector b...